Question

Triangle $ABC$ and $ADC$ are isosceles with $AB=BC$ and $AD=DC$. Point $D$ is inside $\Delta ABC$ $\angle ABC={40}^{\circ }$ and $\angle ADC={140}^{\circ }$. The degree measure of $\angle BAD$ is

• A. $20$
• B. $30$
• C. $40$
• D. $50$

Answer 1

The correct option is D $50$

In triangle $ABC$, $AB=AC$
Thus $\angle BAC=\angle BCA$ (isosceles triangle property)
Sum of angles $={180}^o$
$\angle ABC+\angle BAC+\angle BCA=180$
$2\angle BAC=140$
$\angle BAC={70}^{\circ }$
In triangle $ADC$, $DB=DC$
$\angle DAC=\angle DCA$ (isosceles triangle property)
Sum of angles $={180}^o$
$\angle ADC+\angle DAC+\angle DCA=180$
$2\angle DAC=40$
$\angle DAC={20}^{\circ }$
Thus, $\angle BAD=\angle BAC-\angle DAC$
$\angle BAD=70-20={50}^{\circ }$